# Chapter 17 - Second-Order Differential Equations - 17.4 Series Solutions - 17.4 Exercises - Page 1220: 5

$y=C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}x^n+C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}$

#### Work Step by Step

We have $\Sigma^\infty_{n=0}(n+2) (n+1) C_{n+2} x^{n}+x \Sigma^\infty_{n=1}nC_{n} x^{n-1}+ \Sigma^\infty_{n=0}C_{n}x^{n}=0$ Re-arrange as: $C_{n+2}(n+2)(n+1)+nC_n+C_n=0$ Now, $C_{2} =\dfrac{-C_{0}}{2}$; when $n=0$ and $C_{3} =\dfrac{-C_{1}}{3}$; when $n=1$ After solving, we have $C_{2n} = \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)} \times C_0$ and $C_{2n+1} = \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)} \times C_1$ ...(1) Since, $\Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x}$ Equation (1) becomes: $y= \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)} \times C_0+ \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)} \times C_1$ Hence, we get $y=C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}x^n+C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}$

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